Use the tree diagram to Identify the statement that is not Necessarily True about two independent events, X and Y. Y Р 1-P X Not-X A. q = r. B. (1-q) = q. C. P(X AND Y) = pq. D. P(Y) = pq + (1 − p)r. q 1-q 1-r Not-Y

12. Give the equations that represent each situation. For Independent Events: P(AN B) = P(AU B) =, For Mutually Exclusive Events: P(AN B) = P(AU B) = For Dependent Events: P(AN B) = P(A U B) =

9. By rewriting the formula for the Multiplication Rule, you can write a formula for finding P(A and B) conditional probabilities. The conditional probability of event B occurring, given that event A has occurred, is P(B| A) = P(A) the information below to find the probability that a flight arrives on time given that it departed on time. The probability that an airplane flight departs on time is 0.92. The probability that a flight arrives on time is 0.86. The probability that a flight departs and arrives on time is 0.83. The probability that a flight arrives on time given that it departed on time is (Round to the nearest thousandth as needed.) . Use

Let A and B be two events with P(A and Bc) = 1. Find P(B). Recall that Bc denotes the complement of the event B.

Suppose that  A  and  B  are independent events such that  =PA0.30  and  =PB0.40 . Find  P∩AB  and  P∪AB . (If necessary, consult a list of formulas.) (a) =P∩AB (b) =P∪AB

PLEASE SELECT WHICH OF THE FOLLOWING IS/ARE TRUE AND EXPLAIN. Please answer within 30 minutes.

The events driving a sedan and driving in an attentive style are approximately independent because P(attentive | sedan) ≈ P(attentive).  The events driving a sedan and driving in an attentive style are not independent because P(sedan | attentive) ≈ P(sedan).  The events driving a sedan and driving in an attentive style are approximately independent because P(sedan | attentive) ≈ P(sedan).  The events driving a sedan and driving in an attentive style are not independent because P(attentive | sedan) ≈ P(attentive).